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Completing the Square

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Completing the Square When the Leading Coefficient Doesn't Equal 1
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The area of another parallelogram is given by the equation 3x^2 + 9x - 5 = 0 , where x is the length of the base. What is the length of this base?

Guidance

When there is a number in front of x^2 , it will make completing the square a little more complicated. See how the steps change in Example A.

Example A

Determine the number c that completes the square of 2x^2 - 8x + c .

Solution: In the previous concept, we just added \left(\frac{b}{2}\right)^2 , but that was when a=1 . Now that a \neq 1 , we have to take the value of a into consideration. Let's pull out the the GCF of 2 and 8 first.

2 \left(x^2 - 4x \right)

Now, there is no number in front of x^2 .

\left(\frac{b}{2}\right)^2 = \left(\frac{4}{2}\right)^2 = 4 .

Add this number inside the parenthesis and distribute the 2.

2 \left(x^2 - 4x +4 \right)=2x^2-4x+8

So, c=8 .

Example B

Solve 3x^2-9x+11=0

Solution:

1. Write the polynomial so that x^2 and x are on the left side of the equation and the constants on the right.

3x^2-9x=-11

2. Pull out a from everything on the left side. Even if b is not divisible by a , the coefficient of x^2 needs to be 1 in order to complete the square.

3(x^2-3x+\underline{\;\;\;\;\;\;})=-11

3. Now, complete the square. Determine what number would make a perfect square trinomial.

To do this, divide the x- term by 2 and square that number, or \left(\frac{b}{2}\right)^2 .

\left(\frac{b}{2}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}

4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add {\color{red}a\cdot\left(\frac{b}{2}\right)^2} to keep the equation balanced.

3\left(x^2-3x {\color{red}+\frac{9}{4}}\right)=-11 {\color{red}+\frac{27}{4}}

5. Factor the left side and simplify the right.

3\left(x-\frac{3}{2}\right)^2=-\frac{17}{4}

6. Solve by using square roots.

\left(x-\frac{3}{2}\right)^2 &=-\frac{17}{12}\\x-\frac{3}{2} &=\pm \frac{i\sqrt{17}}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\x &=\frac{3}{2}\pm \frac{\sqrt{51}}{6}i

Be careful with the addition of step 2 and the changes made to step 4. A very common mistake is to add \left(\frac{b}{2}\right)^2 to both sides, without multiplying by a for the right side.

Example C

Solve 4x^2+7x-18=0 .

Solution: Let’s follow the steps from Example B.

1. Write the polynomial so that x^2 and x are on the left side of the equation and the constants on the right.

4x^2-7x=18

2. Pull out a from everything on the left side.

4\left(x^2+\frac{7}{4}x+\underline{\;\;\;\;\;\;}\right)=18

3. Now, complete the square. Find \left(\frac{b}{2}\right)^2 .

\left(\frac{b}{2}\right)^2=\left(\frac{7}{8}\right)^2=\frac{49}{64}

4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add {\color{red}a\cdot\left(\frac{b}{2}\right)^2} to keep the equation balanced.

4\left(x^2+\frac{7}{4}x{\color{red}+\frac{49}{64}}\right)=18{\color{red}+\frac{49}{16}}

5. Factor the left side and simplify the right.

4\left(x+\frac{7}{8}\right)^2=\frac{337}{16}

6. Solve by using square roots.

\left(x+\frac{7}{8}\right)^2 &=\frac{337}{64}\\x+\frac{7}{8} &=\pm \frac{\sqrt{337}}{8}\\x &=-\frac{7}{8} \pm \frac{\sqrt{337}}{8}

Intro Problem Revisit We can't factor 3x^2 + 9x - 5 = 0 , so let's follow the step-by-step process we learned in this lesson.

1. Write the polynomial so that x^2 and x are on the left side of the equation and the constants on the right.

3x^2 + 9x = 5

2. Pull out a from everything on the left side.

3\left(x^2 + 3x+\underline{\;\;\;\;\;\;}\right) = 5

3. Now, complete the square. Find \left(\frac{b}{2}\right)^2 .

\left(\frac{b}{2}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}

4. Add this number to the interior of the parenthesis on the left side. On the right side, you will need to add {\color{red}a\cdot\left(\frac{b}{2}\right)^2} to keep the equation balanced.

3\left(x^2+ 3x{\color{red}+\frac{9}{4}}\right)=5{\color{red}+\frac{27}{4}}

5. Factor the left side and simplify the right.

3\left(x+\frac{3}{2}\right)^2=\frac{47}{4}

6. Solve by using square roots.

\left(x+\frac{3}{2}\right)^2 &=\frac{47}{12}\\x+\frac{3}{2} &=\pm \frac{\sqrt{47}}{\sqrt{12}}\\x &=-\frac{3}{2} \pm \frac{{\sqrt{47}}}{2\sqrt{3}}\\x &=-\frac{3}{2} \pm \frac{{\sqrt{141}}}{6}

However, because x is the length of the parallelogram's base, it must have a positive value. Only x =-\frac{3}{2} + \frac{{\sqrt{141}}}{6} results in a positive value, so the length of the base is x =-\frac{3}{2} + \frac{{\sqrt{141}}}{6} .

Guided Practice

Solve the following quadratic equations by completing the square.

1. 5x^2+29x-6=0

2. 8x^2-32x+4=0

Answers

Use the steps from the examples above to solve for x .

1.

5x^2+29x-6 &=0\\5\left(x^2+\frac{29}{5}x\right) &=6\\5\left(x^2+\frac{29}{5}x+\frac{841}{100}\right) &=6+\frac{841}{20}\\5\left(x+\frac{29}{10}\right)^2 &=\frac{961}{20}\\\left(x+\frac{29}{10}\right)^2 &=\frac{961}{100}\\x+\frac{29}{10} &=\pm \frac{31}{10}\\x &=-\frac{29}{10} \pm \frac{31}{10}\\x &=-6, \frac{1}{5}

2.

8x^2-32x+4 &=0\\8(x^2-4x) &=-4\\8(x^2-4x+4) &=-4+32\\8(x-2)^2 &=28\\(x-2)^2 &=\frac{7}{2}\\x-2 &=\pm \frac{\sqrt{7}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\x &=2\pm \frac{\sqrt{14}}{2}

Practice

Solve the quadratic equations by completing the square.

  1. 6x^2-12x-7=0
  2. -4x^2+24x-100=0
  3. 5x^2-30x+55=0
  4. 2x^2-x-6=0
  5. \frac{1}{2}x^2+7x+8=0
  6. -3x^2+4x+15=0

Solve the following equations by factoring, using square roots, or completing the square.

  1. 4x^2-4x-8=0
  2. 2x^2+9x+7=0
  3. -5(x+4)^2-19=26
  4. 3x^2+30x-5=0
  5. 9x^2-15x-6=0
  6. 10x^2+40x+88=0

Problems 13-15 build off of each other.

  1. Challenge Complete the square for ax^2+bx+c=0 . Follow the steps from Examples A and B. Your final answer should be in terms of a, b, and c .
  2. For the equation 8x^2+6x-5=0 , use the formula you found in #13 to solve for x .
  3. Is the equation in #14 factorable? If so, factor and solve it.
  4. Error Analysis Examine the worked out problem below.

4x^2-48x+11&=0\\4(x^2-12x+\underline{\;\;\;\;\;\;}) &=-11\\4(x^2-12x+36) &=-11+36\\4(x-6)^2 &=25\\(x-6)^2 &=\frac{25}{4}\\x-6 &=\pm \frac{5}{2}\\x &=6\pm \frac{5}{2} \rightarrow \frac{17}{2},\frac{7}{2}

Plug the answers into the original equation to see if they work. If not, find the error and correct it.

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